# How do you solve by completing the square 9x^2+6x=8?

${x}^{2} + \frac{6}{9} x = \frac{8}{9}$
${x}^{2} + \frac{2}{3} x = \frac{8}{9}$
${x}^{2} + \frac{2}{3} x + \frac{1}{9} - \frac{1}{9} = \frac{8}{9}$
${\left(x + \frac{1}{3}\right)}^{2} = \frac{8}{9} + \frac{1}{9}$
$x + \frac{1}{3} = \pm \sqrt{1}$
$x = - \frac{1}{3} \pm 1$
${x}_{1} = \frac{2}{3}$
${x}_{2} = - \frac{4}{3}$